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In Reply to: RE: Voltage Divder posted by deafbykhorns on July 22, 2015 at 11:22:32
"What I meant is the PIC will interpret the decrease in voltage as an increase in current at the cathode."As I said below, that's not correct. A decrease in voltage at the top of the cathode resistor means cathode current has decreased also. The resistor creates an I/V conversion, and the two quantities are directly proportional. Half the voltage means current has also been halved. Double the voltage results from double the current. I think you've confused this mechanism with the increase or decrease in voltage that occurs when the value of the resistor has been changed.
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Edits: 07/22/15 07/22/15Follow Ups:
when I change the cathode resistor to lower the voltage, current goes up.
Is this not the proper way to monitor current at the cathode? Most failures are probably with the tube so wouldn't the higher current reduce the cathode voltage similar to me changing the cathode resistor?
Either way, I'm programming the PIC to look for a range so it monitors the change both ways.4.75v at PIC = 110ma current
4.9v at PIC = 95ma currentMy goal is to protect the output transformer and GM70 if something goes wrong. Should I be looking at another way to monitor current?
Edits: 07/23/15
" wouldn't the higher current reduce the cathode voltage similar to me changing the cathode resistor?"Monitoring cathode voltage has nothing to do with changing the resistor. The resistor is fixed. Ohm's Law tells us how to calculate current from the resistor and voltage. Mathematically, I (current in amps) = E (voltage across the fixed resistor) / R (resistance in Ohms). The PIC needs to be programmed to convert the voltage at the cathode in accordance with this formula. Measure the voltage, divide it by the resistance, and the PIC can store the result as current.
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Edits: 07/23/15 07/23/15
I changed the resistor to simply simulate an increase in current.
For what purpose were you simulating the increase in current? Changing the resistor may not be the correct method.
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Buy Chinese. Bury freedom.
To Monitor voltage at PIC when gm70 pulls enough current at which I want to trigger a relay to protect opt/tube. At that point I can pick a trigger point
You're only confusing yourself by changing the value of the resistor in the sim. When you monitor with the PIC, the resistor won't be changing. Current will be calculated with Ohm's Law using the value of the real, fixed-value resistor and the voltage across it. Those two parameters determine current. Forget the simulation. Forget the vacuum tube. Forget the B+ and the driver and everything else. The only thing you need to know is that current through the resistor equals the voltage across it divided by Ohms. Nothing else matters for purposes of monitoring with the PIC. Put that formula into the PIC along with the value of the actual resistor, and you're done.
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Buy Chinese. Bury freedom.
I don't know why I thought I could change the cathode resistor to simulate current going up (since this resistor wouldn't change unless it failed).
I did the math and found that with the 1.2K cathode resistor and 120ma of current, I would get about 6.2v at the PIC(144v at the top of cathode resistor). I don't do these manual calculations enough so I needed a slap in the face..... Seems I'm getting lazy with computers around!
Now I need to adjust the voltage divider since its a little high for the PIC I'm using.BTW, I did play around with the circuit just to see if I could simulate this as well by simulating a partially shorted winding in the OPT.
I got the same results as the manual calculation. I'll still do the manual calculation to adjust the "window" of current I'm monitoring. It's actually easier than simulating a partial short in an OPT
Thanks again for the help as always
Edits: 07/26/15
It's a common trap to make things harder than they need to be. We all fall into it from time to time, especially using sims. Let me know how the PIC works out. BTW, I might be headed down the same road when I add protection to my 211s. I've found a high voltage opamp that I think can accomplish this with analog technology, but I haven't built a prototype yet. If it proves troublesome, I'll use one of the parts from PICAXE. I built a capacitor surge tester with their PICs a couple years ago, and it worked out really well.
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Buy Chinese. Bury freedom.
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